hdu-2141Can you find it?（二分搜索求和）

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)

Total Submission(s): 24425    Accepted Submission(s): 6186


Problem Description

Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X.

 

Input

There are many cases. Every data case is described as followed: In the first line there are three integers L, N, M, in the second line there are L integers represent the sequence A, in the third line there are N integers represent the sequences B, in the forth
line there are M integers represent the sequence C. In the fifth line there is an integer S represents there are S integers X to be calculated. 1<=L, N, M<=500, 1<=S<=1000. all the integers are 32-integers.

 

Output

For each case, firstly you have to print the case number as the form "Case d:", then for the S queries, you calculate if the formula can be satisfied or not. If satisfied, you print "YES", otherwise print "NO".

 

Sample Input

3 3 3
1 2 3
1 2 3
1 2 3
3
1
4
10

 

Sample Output

Case 1:
NO
YES
NO

 

Author

wangye

 

Source

HDU 2007-11 Programming Contest

 

Recommend

威士忌

 
先把第一组和第二组所以加起来的可能性都保存起来 在从小到大排序 k=sum-c[i]  二分搜索能否找到k 找到即可以输出yes

#include<iostream>
#include<string>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
int a[600],b[600],c[600],sum[360000];
int n,m,l,s;
int se(int num,int k)
{
int l=k-1,r=0;
int mid;
while(l>=r)
{
mid=(l+r)/2;
if(sum[mid]==num)
return 1;
else if(sum[mid]>num)
l=mid-1;
else r=mid+1;
}
return 0;
}
int main()
{
int cnt=1;
while(~scanf("%d%d%d",&l,&n,&m))
{
for(int i=0;i<l;i++)
scanf("%d",&a[i]);
for(int i=0;i<n;i++)
scanf("%d",&b[i]);
for(int i=0;i<m;i++)
scanf("%d",&c[i]);
int k=0;
for(int i=0;i<l;i++)
{
for(int j=0;j<n;j++)
{
sum[k++]=a[i]+b[j];
}
}
sort(sum,sum+k);
scanf("%d", &s);
printf("Case %d:\n", cnt++);
int x;
while(s--)
{
scanf("%d",&x);
int f=0;
for(int i=0;i<m&&f==0;i++)
{
int num=(x-c[i]);
if(se(num,k))
{
f=1;
}
}
if(f)printf("YES\n");
else printf("NO\n");
}
}
return 0;
}